2025 Cambridge IAS-Level Mathematics - Further (9231) 模擬試題連答案詳解

(a)
\(\mathbf{M}^2 = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 3(3) + (-4)(1) & 3(-4) + (-4)(-1) \\ 1(3) + (-1)(1) & 1(-4) + (-1)(-1) \end{pmatrix} = \begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix}\).

\(\mathbf{M}^3 = \mathbf{M}^2 \mathbf{M} = \begin{pmatrix} 5 & -8 \\ 2 & -3 \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \begin{pmatrix} 5(3) + (-8)(1) & 5(-4) + (-8)(-1) \\ 2(3) + (-3)(1) & 2(-4) + (-3)(-1) \end{pmatrix} = \begin{pmatrix} 7 & -12 \\ 3 & -5 \end{pmatrix}\).

(b) Let \(H_n\) be the statement that \(\mathbf{M}^n = \begin{pmatrix} 2n+1 & -4n \\ n & 1-2n \end{pmatrix}\).

**Base case:** For \(n=1\), \(\mathbf{M}^1 = \begin{pmatrix} 2(1)+1 & -4(1) \\ 1 & 1-2(1) \end{pmatrix} = \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix} = \mathbf{M}\). Thus, \(H_1\) is true.

**Inductive step:** Assume that \(H_k\) is true for some positive integer \(k\). That is, \(\mathbf{M}^k = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix}\).

We must show that \(H_{k+1}\) is true, i.e., \(\mathbf{M}^{k+1} = \begin{pmatrix} 2(k+1)+1 & -4(k+1) \\ k+1 & 1-2(k+1) \end{pmatrix} = \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\).

We have:
\(\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M} = \begin{pmatrix} 2k+1 & -4k \\ k & 1-2k \end{pmatrix} \begin{pmatrix} 3 & -4 \\ 1 & -1 \end{pmatrix}\)
\(= \begin{pmatrix} (2k+1)(3) + (-4k)(1) & (2k+1)(-4) + (-4k)(-1) \\ k(3) + (1-2k)(1) & k(-4) + (1-2k)(-1) \end{pmatrix}\)
\(= \begin{pmatrix} 6k+3-4k & -8k-4+4k \\ 3k+1-2k & -4k-1+2k \end{pmatrix}\)
\(= \begin{pmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{pmatrix}\).

This is of the required form for \(n=k+1\). Since \(H_1\) is true, and \(H_k \implies H_{k+1}\), the statement \(H_n\) is true for all positive integers \(n\) by mathematical induction.

(a)
M1: For attempting matrix multiplication for \(\mathbf{M}^2\).
A1: For correct \(\mathbf{M}^2\).
A1: For correct \(\mathbf{M}^3\).

(b)
B1: For verifying the base case \(n=1\) clearly.
M1: For stating the inductive hypothesis for \(n=k\).
M1: For attempting to find \(\mathbf{M}^{k+1} = \mathbf{M}^k \mathbf{M}\) (or vice versa).
A1: For correctly multiplying the matrices.
A1: For simplifying algebraic expressions to the required form.
A1: For a complete and coherent inductive conclusion stating that the result holds for all positive integers.

(a) From the cubic equation \(x^3 - 3x^2 + 2x - 5 = 0\), we have:
\(\sum \alpha = 3\)
\(\sum \alpha\beta = 2\)
\(\alpha\beta\gamma = 5\)
Using the identity:
\(\alpha^2 + \beta^2 + \gamma^2 = (\sum \alpha)^2 - 2\sum \alpha\beta = 3^2 - 2(2) = 9 - 4 = 5\).

(b) Since \(\alpha, \beta, \gamma\) are roots of the cubic equation:
\(\alpha^3 - 3\alpha^2 + 2\alpha - 5 = 0\)
\(\beta^3 - 3\beta^2 + 2\beta - 5 = 0\)
\(\gamma^3 - 3\gamma^2 + 2\gamma - 5 = 0\)
Summing these three equations gives:
\(\sum \alpha^3 - 3\sum \alpha^2 + 2\sum \alpha - 15 = 0\)
Substitute the known values:
\(\sum \alpha^3 - 3(5) + 2(3) - 15 = 0\)
\(\sum \alpha^3 - 15 + 6 - 15 = 0 \implies \sum \alpha^3 = 24\).

(c) Let \(y = \frac{1}{x} \implies x = \frac{1}{y}\).
Substituting \(x = \frac{1}{y}\) into the original equation:
\((\frac{1}{y})^3 - 3(\frac{1}{y})^2 + 2(\frac{1}{y}) - 5 = 0\)
\(\frac{1}{y^3} - \frac{3}{y^2} + \frac{2}{y} - 5 = 0\)
Multiplying by \(y^3\):
\(1 - 3y + 2y^2 - 5y^3 = 0 \implies 5y^3 - 2y^2 + 3y - 1 = 0\).

(a)
B1: Correctly state \(\sum \alpha = 3\) and \(\sum \alpha\beta = 2\).
M1: Use of identity \(\sum \alpha^2 = (\sum \alpha)^2 - 2\sum \alpha\beta\).
A1: Correct value 5.

(b)
M1: State the identity or set up the sum of cubic equations.
M1: Substitute the values of \(\sum \alpha^2\) and \(\sum \alpha\).
A1: Correctly simplify the expression.
A1: Correct value 24.

(c)
M1: Use substitution \(x = \frac{1}{y}\) (or alternative method using sums of roots).
M1: Obtain a fraction-free equation.
A1: Correct polynomial equation in \(y\).
A1: Write final equation with integer coefficients (e.g., \(5y^3 - 2y^2 + 3y - 1 = 0\)).

(a) \(\frac{1}{r^2} - \frac{1}{(r+1)^2} = \frac{(r+1)^2 - r^2}{r^2(r+1)^2} = \frac{r^2+2r+1-r^2}{r^2(r+1)^2} = \frac{2r+1}{r^2(r+1)^2}\) (as required).

(b) Using the difference method:
\(\sum_{r=1}^n \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^n \left( \frac{1}{r^2} - \frac{1}{(r+1)^2} \right)\)
\(= \left(1 - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{9}\right) + \dots + \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)\)
All middle terms cancel, leaving:
\(= 1 - \frac{1}{(n+1)^2}\).

(c) As \(n \to \infty\), \(\frac{1}{(n+1)^2} \to 0\).
Therefore, \(\sum_{r=1}^\infty \frac{2r+1}{r^2(r+1)^2} = \lim_{n\to\infty} \left( 1 - \frac{1}{(n+1)^2} \right) = 1\).

(d) We want \(\sum_{r=N}^\infty \frac{2r+1}{r^2(r+1)^2} < 0.005\).
\(\sum_{r=N}^\infty \frac{2r+1}{r^2(r+1)^2} = \sum_{r=1}^\infty \frac{2r+1}{r^2(r+1)^2} - \sum_{r=1}^{N-1} \frac{2r+1}{r^2(r+1)^2}\)
\(= 1 - \left( 1 - \frac{1}{N^2} \right) = \frac{1}{N^2}\).
We set \(\frac{1}{N^2} < 0.005 \implies N^2 > 200 \implies N > \sqrt{200} \approx 14.14\).
Since \(N\) is an integer, the smallest value of \(N\) is 15.

(a)
M1: For putting fractions over a common denominator.
A1: Correct algebraic expansion leading to numerator \(2r+1\).

(b)
M1: Expressing the series using the result from part (a).
A1: Writing down the first few terms showing the differences.
A1: Indicating cancellation clearly.
A1: Obtaining the correct simplified sum \(1 - \frac{1}{(n+1)^2}\).

(c)
M1: Taking the limit of the sum as \(n \to \infty\).
A1: Deducting the correct limit value of 1.

(d)
M1: Expressing the sum from \(N\) to infinity as \(\frac{1}{N^2}\).
A1: Setting up and solving the inequality \(N^2 > 200\).
A1: Concluding the integer value \(N = 15\).

(a) A vector perpendicular to both lines is the cross product of \(\mathbf{d}_1\) and \(\mathbf{d}_2\):
\(\mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -1 \\ 2 & -1 & 3 \end{vmatrix}\)
\(= \mathbf{i}(6 - 1) - \mathbf{j}(3 + 2) + \mathbf{k}(-1 - 4) = 5\mathbf{i} - 5\mathbf{j} - 5\mathbf{k}\).
We can scale this to simplify the normal vector to: \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\).

(b) The vector \(\vec{AB}\) is:
\(\vec{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 2 \\ 3 \\ 0 \end{pmatrix} - \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 1 \\ 4 \\ -2 \end{pmatrix}\).
The shortest distance \(d\) is the projection of \(\vec{AB}\) onto \(\mathbf{n}_0\):
\(d = \frac{| \vec{AB} \cdot \mathbf{n}_0 |}{|\mathbf{n}_0|}\)
\(\vec{AB} \times \mathbf{n}_0 = (1)(1) + (4)(-1) + (-2)(-1) = 1 - 4 + 2 = -1\).
\(|\mathbf{n}_0| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}\).
Thus, \(d = \frac{|-1|}{\sqrt{3}} = \frac{\sqrt{3}}{3} \approx 0.577\).

(c) The plane has normal vector \(\mathbf{n}_0 = \mathbf{i} - \mathbf{j} - \mathbf{k}\) and contains the point \(A(1, -1, 2)\).
Its equation is:
\(1(x - 1) - 1(y + 1) - 1(z - 2) = 0\)
\(x - y - z = 0\).

(a)
M1: For attempting the cross product of \(\mathbf{d}_1\) and \(\mathbf{d}_2\).
A1: For evaluating at least two components correctly.
A1: Correct perpendicular vector (any scalar multiple of \(\mathbf{i} - \mathbf{j} - \mathbf{k}\)).

(b)
B1: For finding the vector \(\vec{AB} = \mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\).
M1: Using the formula for shortest distance between two skew lines.
A1: Evaluating \(|\vec{AB} \cdot \mathbf{n}_0|\) correctly.
A1: Correctly finding the magnitude of the normal vector.
A1: Correct distance \(\frac{1}{\sqrt{3}}\) or \(\frac{\sqrt{3}}{3}\).

(c)
M1: Recognizing the plane's normal is \(\mathbf{n}_0\) and contains \(A\).
A1: Setting up \(\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n}\).
A1: Correct Cartesian equation \(x - y - z = 0\).

(a) Let \(\begin{pmatrix} x \\ y \end{pmatrix}\) be an invariant point. Then:
\(\begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} x \\ y \end{pmatrix}\)
This gives:
\(3x + 2y = x \implies 2x + 2y = 0 \implies x + y = 0\)
\(x + 4y = y \implies x + 3y = 0\)
Solving simultaneously:
Subtracting the first from the second gives \(2y = 0 \implies y = 0\), which leads to \(x = 0\).
Thus, the origin \((0,0)\) is the unique invariant point.

(b) Let the invariant line be \(y = mx\).
Let \(\begin{pmatrix} X \\ Y \end{pmatrix} = \begin{pmatrix} 3 & 2 \\ 1 & 4 \end{pmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} 3x + 2y \\ x + 4y \end{pmatrix}\).
Since a point on \(y=mx\) maps to another point on \(Y=mX\):
\(x + 4(mx) = m(3x + 2mx)\)
Since this must hold for all \(x \neq 0\), we divide by \(x\):
\(1 + 4m = m(3 + 2m) \implies 2m^2 - m - 1 = 0\)
\((2m + 1)(m - 1) = 0 \implies m = 1\) or \(m = -\frac{1}{2}\).
Thus, the invariant lines are \(y = x\) and \(y = -\frac{1}{2}x\).

(c) The determinant of \(\mathbf{M}\) is \(3(4) - 2(1) = 10\).
The inverse is \(\mathbf{M}^{-1} = \frac{1}{10} \begin{pmatrix} 4 & -2 \\ -1 & 3 \end{pmatrix}\).
Thus:
\(x = \frac{4X - 2Y}{10}\) and \(y = \frac{-X + 3Y}{10}\).
Substituting into \(y = 2x + 1\):
\(\frac{-X + 3Y}{10} = 2\left(\frac{4X - 2Y}{10}\right) + 1\)
Multiply by 10:
\(-X + 3Y = 2(4X - 2Y) + 10\)
\(-X + 3Y = 8X - 4Y + 10 \implies 9X - 7Y + 10 = 0\).
Thus, the equation of the image line is \(9x - 7y + 10 = 0\).

(a)
M1: For setting up the matrix equation \(\mathbf{M}\mathbf{x} = \mathbf{x}\).
A1: For obtaining both simultaneous equations.
A1: For solving to show that the only solution is \((0, 0)\).

(b)
M1: Substituting \(y=mx\) and \(Y=mX\) into the transformation equations.
A1: Obtaining the quadratic equation in \(m\), e.g., \(2m^2 - m - 1 = 0\).
M1: Factoring or solving the quadratic equation.
A1: Finding the two correct gradients \(m = 1\) and \(m = -1/2\).
A1: Correct equations of both lines.

(c)
M1: Finding the inverse matrix \(\mathbf{M}^{-1}\).
A1: Substituting \(x\) and \(y\) in terms of \(X\) and \(Y\) into the line equation.
A1: Correct final equation (e.g., \(9x - 7y + 10 = 0\) or \(y = \frac{9}{7}x + \frac{10}{7}\)).

(a) The curve is a cardioid restricted to the upper half-plane since \(0 \le \theta \le \pi\).
- At \(\theta = 0\), \(r = 0\) (the pole).
- At \(\theta = \frac{\pi}{2}\), \(r = 2(1 - 0) = 2\).
- At \(\theta = \pi\), \(r = 2(1 - (-1)) = 4\).
The sketch is a cardioid-like arc starting at the pole, curving through \((2, \frac{\pi}{2})\), and ending at \((4, \pi)\).

(b) The area \(A\) is:
\(A = \frac{1}{2} \int_{0}^{\pi/2} r^2 \mathrm{d}\theta = \frac{1}{2} \int_{0}^{\pi/2} 4(1 - \cos \theta)^2 \mathrm{d}\theta\)
\(= 2 \int_{0}^{\pi/2} (1 - 2\cos \theta + \cos^2 \theta) \mathrm{d}\theta\)
Using \(\cos^2 \theta = \frac{1 + \cos 2\theta}{2}\):
\(= 2 \int_{0}^{\pi/2} \left( \frac{3}{2} - 2\cos \theta + \frac{1}{2}\cos 2\theta \right) \mathrm{d}\theta\)
\(= 2 \left[ \frac{3}{2}\theta - 2\sin \theta + \frac{1}{4}\sin 2\theta \right]_0^{\pi/2}\)
\(= 2 \left( \left( \frac{3\pi}{4} - 2 + 0 \right) - (0) \right) = \frac{3\pi}{2} - 4\).

(c) The distance from the pole is \(r = 2(1 - \cos \theta)\).
Since \(-1 \le \cos \theta \le 1\), the maximum value of \(r\) occurs when \(\cos \theta = -1\), which is at \(\theta = \pi\).
At \(\theta = \pi\), \(r = 2(1 - (-1)) = 4\).

(a)
B1: Starts at the pole with correct slope.
B1: Passes through \((2, \frac{\pi}{2})\) and \((4, \pi)\).
B1: Correct shape (smooth upper half cardioid) with no loops.

(b)
M1: Use of correct area formula \(\frac{1}{2}\int r^2 \mathrm{d}\theta\) with limits \(0\) to \(\frac{\pi}{2}\).
M1: Attempting to square \(1 - \cos \theta\).
M1: Using double angle identity for \(\cos^2 \theta\).
A1: Correct integration.
A1: Correct exact value \(\frac{3̄\pi}{2} - 4\).

(c)
M1: Recognizing that maximum distance corresponds to minimum of \(\cos \theta\).
A1: Correct maximum distance of 4.
A1: Correct value \(\theta = \pi\).

(a) The denominator of \(y\) is \(x^2 - 1 = (x - 1)(x + 1)\). Setting the denominator to 0 gives \(x = 1\) and \(x = -1\). Since the numerator \(2x^2 + 3x - 2 = (2x - 1)(x + 2)\) has no common factors with the denominator, the vertical asymptotes are \(x = 1\) and \(x = -1\).

For the horizontal asymptote:
\(\lim_{x \to \pm \infty} \frac{2x^2+3x-2}{x^2-1} = 2\).
Thus, the horizontal asymptote is \(y = 2\).

(b)
- y-intercept: set \(x = 0 \implies y = \frac{-2}{-1} = 2\). So the intercept is \((0, 2)\).
- x-intercepts: set \(y = 0 \implies 2x^2 + 3x - 2 = 0 \implies (2x - 1)(x + 2) = 0\), which gives \(x = \frac{1}{2}\) or \(x = -2\). So the intercepts are \((\frac{1}{2}, 0)\) and \((-2, 0)\).

(c) Using the quotient rule:
\(\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(4x + 3)(x^2 - 1) - (2x^2 + 3x - 2)(2x)}{(x^2 - 1)^2}\)
\(= \frac{(4x^3 - 4x + 3x^2 - 3) - (4x^3 + 6x^2 - 4x)}{(x^2 - 1)^2}\)
\(= \frac{-3x^2 - 3}{(x^2 - 1)^2} = -\frac{3(x^2 + 1)}{(x^2 - 1)^2}\).
Since \(3(x^2+1) > 0\) and \((x^2-1)^2 > 0\) for all \(x \neq \pm 1\), we have \(\frac{\mathrm{d}y}{\mathrm{d}x} < 0\) for all points on \(C\).

(d) To sketch \(C\):
- Draw asymptotes \(x = -1\), \(x = 1\), and \(y = 2\).
- Plot points \((0, 2)\), \((\frac{1}{2}, 0)\), and \((-2, 0)\).
- Draw three branches of the curve showing decreasing behavior in each domain segment.

(a)
B1: Identify both vertical asymptotes \(x = 1\) and \(x = -1\).
M1: For attempting the limit as \(x \to \infty\).
A1: Correct horizontal asymptote \(y = 2\).

(b)
B1: Correct y-intercept \((0, 2)\).
M1: Set numerator to 0 and solve.
A1: Correct x-intercepts \((\frac{1}{2}, 0)\) and \((-2, 0)\).

(c)
M1: Attempting differentiation using the quotient rule.
A1: Correctly simplifying to obtain \(-\frac{3(x^2+1)}{(x^2-1)^2}\) and concluding.

(d)
B1: All asymptotes correctly drawn and labeled.
B1: Three branches correctly plotted with decreasing shape.
B1: All three coordinate axis intercepts correctly located.

First, find the complementary function (CF) by solving the auxiliary equation:
\[ m^2 + 4m + 13 = 0 \implies m = -2 \pm 3i \]
Thus, the CF is:
\[ y_c = e^{-2x} (A \cos 3x + B \sin 3x) \]
Next, find the particular integral (PI) by trying a linear function:
\[ y_p = px + q \implies y_p' = p, \ y_p'' = 0 \]
Substituting into the differential equation:
\[ 0 + 4p + 13(px + q) = 169x \]
Comparing coefficients of \( x \):
\[ 13p = 169 \implies p = 13 \]
Comparing constant terms:
\[ 4p + 13q = 0 \implies 52 + 13q = 0 \implies q = -4 \]
So, the general solution is:
\[ y = e^{-2x} (A \cos 3x + B \sin 3x) + 13x - 4 \]
Apply initial condition \( y(0) = 0 \):
\[ 0 = A - 4 \implies A = 4 \]
Differentiate the general solution:
\[ \frac{dy}{dx} = -2e^{-2x} (A \cos 3x + B \sin 3x) + e^{-2x} (-3A \sin 3x + 3B \cos 3x) + 13 \]
Apply initial condition \( \frac{dy}{dx}(0) = 0 \):
\[ 0 = -2A + 3B + 13 \]
Since \( A = 4 \):
\[ 0 = -8 + 3B + 13 \implies 3B = -5 \implies B = -\frac{5}{3} \]
Thus, the particular solution is:
\[ y = e^{-2x} \left(4 \cos 3x - \frac{5}{3} \sin 3x\right) + 13x - 4 \]

M1: Set up and solve auxiliary equation to find complex roots.
A1: Correct complementary function.
M1: Set up the correct form of the PI and substitute into the differential equation.
A1: Correct particular integral \( y_p = 13x - 4 \).
M1: Apply initial condition \( y(0) = 0 \) to the general solution.
A1: Find \( A = 4 \).
M1: Correctly differentiate the general solution and apply the second initial condition.
A1: Find \( B = -\frac{5}{3} \).
A1: State the final particular solution correctly.

Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal elements:
\[ \lambda_1 = 2, \quad \lambda_2 = 3, \quad \lambda_3 = 4 \]
To find the eigenvectors:
For \( \lambda_1 = 2 \):
\[ (\mathbf{A} - 2\mathbf{I})\mathbf{v} = \begin{pmatrix} 0 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This gives \( z = 0 \), \( y = 0 \), and \( x \) is free. Hence, a unit eigenvector is:
\[ \mathbf{v}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \]

For \( \lambda_2 = 3 \):
\[ (\mathbf{A} - 3\mathbf{I})\mathbf{v} = \begin{pmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This gives \( z = 0 \) and \( x = y \). An eigenvector is \( \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \). Normalising it to unit length:
\[ \mathbf{v}_2 = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \]

For \( \lambda_3 = 4 \):
\[ (\mathbf{A} - 4\mathbf{I})\mathbf{v} = \begin{pmatrix} -2 & 1 & 1 \\ 0 & -1 & 2 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix} \]
This gives \( y = 2z \) and \( -2x + y + z = 0 \implies -2x + 3z = 0 \implies x = 1.5z \). Choosing \( z = 2 \), we have \( y = 4 \) and \( x = 3 \). Hence, an eigenvector is \( \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix} \). Normalising it to unit length (length is \(\sqrt{9+16+4} = \sqrt{29}\)):
\[ \mathbf{v}_3 = \frac{1}{\sqrt{29}} \begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix} = \begin{pmatrix} 3/\sqrt{29} \\ 4/\sqrt{29} \\ 2/\sqrt{29} \end{pmatrix} \]

B1: Identify the eigenvalues as 2, 3, 4.
M1: Write down the characteristic system for \(\lambda = 2\).
A1: Obtain the correct unit eigenvector \(\begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}\).
M1: Write down the characteristic system for \(\lambda = 3\).
A1: Obtain the correct unit eigenvector \(\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}\).
M1: Write down the characteristic system for \(\lambda = 4\).
A1: Obtain a correct eigenvector for \(\lambda = 4\) (e.g., \(\begin{pmatrix} 3 \\ 4 \\ 2 \end{pmatrix}\)).
A2: Normalise and state the correct unit eigenvector for \(\lambda = 4\).

Let \( f(x) = \ln(1 + \sin x) \).
At \( x = 0 \): \( f(0) = \ln(1) = 0 \).

First derivative:
\[ f'(x) = \frac{\cos x}{1 + \sin x} \]
At \( x = 0 \): \( f'(0) = \frac{1}{1} = 1 \).

Second derivative (using quotient rule):
\[ f''(x) = \frac{-\sin x(1 + \sin x) - \cos x(\cos x)}{(1 + \sin x)^2} = \frac{-\sin x - (\sin^2 x + \cos^2 x)}{(1 + \sin x)^2} = \frac{-1 - \sin x}{(1 + \sin x)^2} = -\frac{1}{1 + \sin x} \]
At \( x = 0 \): \( f''(0) = -\frac{1}{1} = -1 \).

Third derivative:
Rewriting \( f''(x) = -(1 + \sin x)^{-1} \):
\[ f'''(x) = (1 + \sin x)^{-2} \cos x = \frac{\cos x}{(1 + \sin x)^2} \]
At \( x = 0 \): \( f'''(0) = \frac{1}{1^2} = 1 \).

Substitute the values into the Maclaurin's series formula:
\[ f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \dots \]
\[ f(x) = 0 + 1x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + \dots = x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \]

M1: Correctly differentiate to find \( f'(x) \).
A1: Show \( f'(0) = 1 \).
M1: Correctly differentiate to find \( f''(x) \).
A1: Simplify \( f''(x) \) to \( -\frac{1}{1 + \sin x} \) or equivalent.
A1: Show \( f''(0) = -1 \).
M1: Correctly differentiate to find \( f'''(x) \).
A1: Show \( f'''(0) = 1 \).
M1: Apply the Maclaurin formula with the calculated derivatives.
A1: Obtain the correct expansion up to \( x^3 \).

(a) Apply integration by parts with \( u = x^n \) and \( dv = \sin x \, dx \):
\[ du = n x^{n-1} \, dx, \quad v = -\cos x \]
\[ I_n = \left[-x^n \cos x\right]_0^{\frac{\pi}{2}} + n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx \]
Since \( \cos(\frac{\pi}{2}) = 0 \) and \( x^n = 0 \) at \( x = 0 \) (for \( n \ge 2 \)), the boundary term is zero. Thus:
\[ I_n = n \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx \]
Apply integration by parts again with \( u = x^{n-1} \) and \( dv = \cos x \, dx \):
\[ du = (n-1)x^{n-2} \, dx, \quad v = \sin x \]
\[ \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx = \left[x^{n-1} \sin x\right]_0^{\frac{\pi}{2}} - (n-1) \int_0^{\frac{\pi}{2}} x^{n-2} \sin x \, dx \]
At the upper limit \( x = \frac{\pi}{2} \), the term is \( \left(\frac{\pi}{2}\right)^{n-1} \), and at \( x = 0 \), it is 0. So:
\[ \int_0^{\frac{\pi}{2}} x^{n-1} \cos x \, dx = \left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2} \]
Substituting this back gives:
\[ I_n = n \left[ \left(\frac{\pi}{2}\right)^{n-1} - (n-1)I_{n-2} \right] = n \left(\frac{\pi}{2}\right)^{n-1} - n(n-1)I_{n-2} \]

(b) For \( n = 3 \):
\[ I_3 = 3 \left(\frac{\pi}{2}\right)^2 - 6 I_1 = \frac{3\pi^2}{4} - 6 I_1 \]
We find \( I_1 \):
\[ I_1 = \int_0^{\frac{\pi}{2}} x \sin x \, dx = \left[-x \cos x\right]_0^{\frac{\pi}{2}} + \int_0^{\frac{\pi}{2}} \cos x \, dx = 0 + \left[\sin x\right]_0^{\frac{\pi}{2}} = 1 \]
So:
\[ I_3 = \frac{3\pi^2}{4} - 6(1) = \frac{3\pi^2}{4} - 6 \]

Part (a):
M1: Apply integration by parts to \( I_n \).
A1: Obtain the correct expression in terms of \( \int x^{n-1} \cos x \, dx \) with boundary terms vanishing.
M1: Apply integration by parts a second time.
A1: Correctly evaluate the boundary term \( \left(\frac{\pi}{2}\right)^{n-1} \).
A1: Combine the results to complete the proof of the reduction formula.

Part (b):
M1: Substitute \( n = 3 \) into the reduction formula.
M1: Calculate \( I_1 \) correctly using integration by parts.
A1: Find \( I_1 = 1 \).
A1: State the final exact answer for \( I_3 \) correctly.

Express \(\sinh x\) and \(\cosh x\) in terms of exponentials:
\[ 6 \left(\frac{e^x - e^{-x}}{2}\right) - \left(\frac{e^x + e^{-x}}{2}\right) = 4 \]
Multiply the entire equation by 2 to clear the denominators:
\[ 6(e^x - e^{-x}) - (e^x + e^{-x}) = 8 \]
Simplify the terms:
\[ 6e^x - 6e^{-x} - e^x - e^{-x} = 8 \]
\[ 5e^x - 7e^{-x} = 8 \]
Multiply by \( e^x \) to form a quadratic equation:
\[ 5e^{2x} - 8e^x - 7 = 0 \]
Let \( u = e^x \). The equation becomes:
\[ 5u^2 - 8u - 7 = 0 \]
Solve using the quadratic formula:
\[ u = \frac{8 \pm \sqrt{(-8)^2 - 4(5)(-7)}}{10} = \frac{8 \pm \sqrt{64 + 140}}{10} = \frac{8 \pm \sqrt{204}}{10} = \frac{4 \pm \sqrt{51}}{5} \]
Since \( e^x > 0 \) for all real \( x \), we reject the negative root (since \( 4 - \sqrt{51} < 0 \)).
Thus, we have:
\[ e^x = \frac{4 + \sqrt{51}}{5} \]
Take the natural logarithm of both sides:
\[ x = \ln\left(\frac{4 + \sqrt{51}}{5}\right) \]

M1: Express \(\sinh x\) and \(\cosh x\) in exponential form.
A1: Write simplified linear combination of \(e^x\) and \(e^{-x}\).
M1: Multiply by \(e^x\) to form a quadratic in \(e^x\).
A1: Correct quadratic equation \( 5e^{2x} - 8e^x - 7 = 0 \).
M1: Solve the quadratic equation correctly.
A1: Obtain the roots \( \frac{4 \pm \sqrt{51}}{5} \).
M1: Provide valid justification for rejecting the negative root.
A2: Give the final exact answer in the required logarithmic form.

(a) By de Moivre's theorem:
\[ \cos(5\theta) + i\sin(5\theta) = (\cos\theta + i\sin\theta)^5 \]
Expand using the binomial theorem:
\[ (\cos\theta + i\sin\theta)^5 = \cos^5\theta + 5i\cos^4\theta\sin\theta - 10\cos^3\theta\sin^2\theta - 10i\cos^2\theta\sin^3\theta + 5\cos\theta\sin^4\theta + i\sin^5\theta \]
Equate imaginary parts:
\[ \sin(5\theta) = 5\cos^4\theta\sin\theta - 10\cos^2\theta\sin^3\theta + \sin^5\theta \]
Substitute \( \cos^2\theta = 1 - \sin^2\theta \):
\[ \sin(5\theta) = 5(1 - \sin^2\theta)^2\sin\theta - 10(1 - \sin^2\theta)\sin^3\theta + \sin^5\theta \]
\[ = 5(1 - 2\sin^2\theta + \sin^4\theta)\sin\theta - 10(\sin^3\theta - \sin^5\theta) + \sin^5\theta \]
\[ = 5\sin\theta - 10\sin^3\theta + 5\sin^5\theta - 10\sin^3\theta + 10\sin^5\theta + \sin^5\theta \]
\[ = 16\sin^5\theta - 20\sin^3\theta + 5\sin\theta \]

(b) The equation is equivalent to:
\[ \sin(5\theta) = 0 \]
For \( 0 \le \theta < \pi \), we have \( 0 \le 5\theta < 5\pi \).
Therefore, the solutions for \( 5\theta \) are:
\[ 5\theta = 0, \pi, 2\pi, 3\pi, 4\pi \]
Divide by 5:
\[ \theta = 0, \frac{\pi}{5}, \frac{2\pi}{5}, \frac{3\pi}{5}, \frac{4\pi}{5} \]

Part (a):
M1: Set up de Moivre's equation for \( \sin(5\theta) \).
A1: Correct binomial expansion.
M1: Equate imaginary parts of both sides.
A1: Correctly substitute \( \cos^2\theta = 1 - \sin^2\theta \).
A1: Simplify and show the given identity clearly.

Part (b):
M1: Recognize the equation is \( \sin(5\theta) = 0 \).
M1: Find the correct range of \( 5\theta \).
A1: List correct multiple angles.
A1: State all five correct solutions for \( \theta \).

First, write the differential equation in standard linear form by dividing by \( x \):
\[ \frac{dy}{dx} - \frac{3}{x}y = x^4 \cos x \]
The integrating factor is:
\[ I(x) = e^{\int -\frac{3}{x} \, dx} = e^{-3\ln x} = x^{-3} = \frac{1}{x^3} \]
Multiply both sides of the standard form ODE by the integrating factor:
\[ \frac{1}{x^3} \frac{dy}{dx} - \frac{3}{x^4}y = x \cos x \]
This can be written as:
\[ \frac{d}{dx} \left( \frac{y}{x^3} \right) = x \cos x \]
Integrate both sides with respect to \( x \):
\[ \frac{y}{x^3} = \int x \cos x \, dx \]
Using integration by parts with \( u = x \) and \( dv = \cos x \, dx \):
\[ \int x \cos x \, dx = x \sin x - \int \sin x \, dx = x \sin x + \cos x + C \]
Therefore:
\[ \frac{y}{x^3} = x \sin x + \cos x + C \]
Multiply by \( x^3 \) to find the general solution:
\[ y = x^4 \sin x + x^3 \cos x + C x^3 \]

M1: Write the differential equation in standard form.
M1: Determine the correct integrating factor \( I(x) = x^{-3} \).
A1: Multiply the ODE correctly by the integrating factor.
M1: Rewrite the LHS as a single derivative \( \frac{d}{dx}(y x^{-3}) \).
M1: Integrate the RHS using integration by parts.
A1: Correct integration to get \( x \sin x + \cos x \).
A1: Include constant of integration \( C \).
A2: Solve explicitly for \( y \) to obtain the correct final general solution.

Let \( x = \sinh \theta \). Then \( dx = \cosh \theta \, d\theta \).
The term in the denominator becomes:
\[ \sqrt{1 + x^2} = \sqrt{1 + \sinh^2 \theta} = \cosh \theta \]
Change the limits of integration:
When \( x = 0 \), \( \theta = 0 \).
When \( x = 1 \), \( \theta = \sinh^{-1}(1) = \ln(1 + \sqrt{2}) \). Let \( \alpha = \ln(1 + \sqrt{2}) \).

Substitute into the integral:
\[ \int_{0}^{1} \frac{x^2}{\sqrt{1 + x^2}} \, dx = \int_{0}^{\alpha} \frac{\sinh^2 \theta}{\cosh \theta} \cosh \theta \, d\theta = \int_{0}^{\alpha} \sinh^2 \theta \, d\theta \]
Use the identity \( \sinh^2 \theta = \frac{\cosh(2\theta) - 1}{2} \):
\[ \int_{0}^{\alpha} \sinh^2 \theta \, d\theta = \frac{1}{2} \int_{0}^{\alpha} (\cosh(2\theta) - 1) \, d\theta = \frac{1}{2} \left[ \frac{1}{2}\sinh(2\theta) - \theta \right]_0^{\alpha} \]
Using \( \sinh(2\theta) = 2\sinh\theta\cosh\theta \), this simplifies to:
\[ \frac{1}{2} [\sinh\theta\cosh\theta - \theta]_0^{\alpha} = \frac{1}{2} (\sinh\alpha\cosh\alpha - \alpha) \]
Since \( \sinh\alpha = 1 \), we have \( \cosh\alpha = \sqrt{1 + \sinh^2\alpha} = \sqrt{2} \).
Substituting these values:
\[ \frac{1}{2} \left( 1 \cdot \sqrt{2} - \ln(1 + \sqrt{2}) \right) = \frac{1}{2}\sqrt{2} - \frac{1}{2}\ln(1 + \sqrt{2}) \]

M1: Substitute \( x = \sinh\theta \) and correctly differentiate to find \( dx \).
A1: Successfully substitute into the integral to obtain \( \int \sinh^2 \theta \, d\theta \).
M1: Change limits correctly using the logarithmic definition of \( \sinh^{-1}(1) \).
M1: Use hyperbolic double-angle identity to integrate.
A1: Correct integration to \( \frac{1}{4}\sinh(2\theta) - \frac{1}{2}\theta \).
M1: Use \( \sinh(2\theta) = 2\sinh\theta\cosh\theta \) to simplify evaluation.
A1: Determine \( \cosh\alpha = \sqrt{2} \).
A2: State final answer in the exact requested form.